Answer:
a)100
b) (0, -11 + √(84 - (z - 6)^2), z) and (0, -11 - √(84 - (z - 6)^2), z)
c)(2 + √(64 - (y + 11)^2), y, 0) and (2 - √(64 - (y + 11)^2), y, 0)
Explanation:
a)
An equation of a sphere with center (a, b, c) and radius r has the form:
(x- a)^2 + (y - b)^2 + (z - c)^2 = r^2
We are given the center of the sphere as (2, -11, 6) and the radius as 10. Substituting these values into the equation, we get:
(x - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100
So the equation of the sphere is (x - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100.
b) To find the intersection of the sphere with the yz-plane (x = 0), we substitute x = 0 into the equation of the sphere:
(0 - 2)^2 + (y + 11)^2 + (z - 6)^2 = 100
(y + 11)^2 + (z - 6)^2 = 84
This is the equation of a circle with center (0, -11, 6) and radius √84. To find the point(s) of intersection with the yz-plane, we set x = 0 in the equation of the sphere and solve for y and z:
(y + 11)^2 + (z - 6)^2 = 84
y = -11 ± √(84 - (z - 6)^2)
So the two points of intersection with the yz-plane are
(0, -11 + √(84 - (z - 6)^2), z) and (0, -11 - √(84 - (z - 6)^2), z).
c)
To find the intersection of the sphere with the xy-plane (z = 0), we substitute z = 0 into the equation of the sphere:
(x - 2)^2 + (y + 11)^2 + (0 - 6)^2 = 100
(x - 2)^2 + (y + 11)^2 = 64
This is the equation of a circle with center (2, -11) and radius 8. To find the point(s) of intersection with the xy-plane, we set z = 0 in the equation of the sphere and solve for x and y:
(x - 2)^2 + (y + 11)^2 = 64
x = 2 ± √(64 - (y + 11)^2)So the two points of intersection with the xy-plane are (2 + √(64 - (y + 11)^2), y, 0) and (2 - √(64 - (y + 11)^2), y, 0).